Answer
a) Please see below.
b) No.
Work Step by Step
a)
Let $y=x+1$, and $x$ is a multiple of three. Thus, $y$ is one more than a multiple of three.
$y^2 = (x+1)(x+1)= x^2+2x+1$
$x^2$ and $2x$ are multiples of three. The remaining number is 1. Thus, if an integer is one more than a multiple of three, its square is also one more than a multiple of three.
b)
Let $y = x+2$, and $x$ is a multiple of three. Thus, $y$ is two more than a multiple of three.
$y^2 = (x+2)(x+2) = x^2+4x+4 = x^2+ 4x+3+1$
$x^2$, $4x$, and $3$ are multiples of three. The remaining number is 1. Thus, if an integer is two more than a multiple of three, its square is not two more than a multiple of three.
